How do you find the roots, real and imaginary, of $y=-(x -4 )^2-12x+16$ using the quadratic formula?

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Answer 1 · 2017-12-21T01:23:53

The two real roots are:

$x_1\approx 18.24621\quad,quad x_2\approx 1.75379$

First let’s simplify the first parenthetical term, $-(x-4)^2$ , by distributing the negative sign and FOILing it.

$-(x-4)^2$

$color(magenta)(\implies) (-x+4)(-x+4)$

$color(magenta)(\implies) x^2-4x-4x+16$

$color(magenta)(\implies) x^2-8x+16$

Add that to the rest of the original expression:

$color(red)(x^2)-color(blue)(8x)+color(green)(16)-color(blue)(12x)+color(green)(16)$

$color(magenta)(\implies) color(red)(x^2)-color(blue)(20x)+color(green)(32)$

This is in standard form, $ax^2+bx+c$ , so we can solve using the quadratic formula. Here,

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$color(magenta)(\implies)x=\frac{20\pm\sqrt{(-20)^2-4(1)(32)}}{2(10)}$

$color(magenta)(\implies)x=\frac{20\pm\sqrt{400-128}}{2}$

$color(magenta)(\implies)x=\frac{20\pm\sqrt{272}}{2}$

$color(magenta)(\implies)x=\frac{20\pm 4\sqrt{17}}{2}$

$color(magenta)(\implies) x=10\pm 2\sqrt{17}$

$color(magenta)(\therefore) x_1\approx 18.24621\quad,\quad x_2\approx 1.75379$

How do you find the roots, real and imaginary, of y=-(x -4 )^2-12x+16y=-(x -4 )^2-12x+16 using the quadratic formula?